+3 votes
in Class 11 by kratos

Two heater A and B are in parallel across supply voltage V. Heater A produces 500 kcal in 200 min. and B produces 1000 kcal in 10 min. The resistance of A is 10 ohm. What is the resistance of B ? If the same heaters are connected in series across the voltage V, how much heat will be produced in kcal in 5 min ?

1 Answer

+1 vote
by kratos
 
Best answer

Heat produced = V2t/JR kcal

For heater A, 500 = (V2 x (20 x 60))/(R x J) ......(i)

For heater B, 1000 = (V2 x (10 x 60))/(R x J) .....(ii)

From Eq. (i) and (ii), we get, R = 2.5 Ω.

When the two heaters are connected in series, let H be the amount of heat produced in kcal.

Since combined resistance is (10 + 2.5) = 12.5 Ω, hence

H = (V2 x (2 x 60))/(12.5 x J) .....(iii)

Dividing Eq. (iii) by Eq. (i), we have H = 100 kcal.

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